Proof 2

Assume in homogeneous coordinates $A = (1, 0, 0),$ $B = (0, 1, 0),$ $C = (0, 0, 1),$ $P = (a,b,c),$ and the line at infinity given by $x+y+z=0.$ Then, for example, $AB,$ having equation $z=0,$ intersects the line at infinity at point $(1,-1,0)$ and the same is true for the line through $P(a,b,c)$ parellel to $AB.$ This line then meets $BC$ in $(a,b,c)-a(1,-1,0)=(0,a+b,c).$ This is point $C_a$ in the diagram: $C_{a}=(0,a+b,c).$

Similarly, $B_{a} = (0, b, a+c),$ $A_{c} = (a, b+c, 0),$ $B_{c} = (a+c, b, 0),$ $A_{b} = (a, 0, b+c)$ and $C_{b} = (a+b, 0, c).$ To test that they are on a conic, the obvious way is to use Pascal's theorem. Line $B_{a}C_{a}$ has the equation $a=0;$ line $B_{c}C_{b}$ $tB_{c}+sC_{b}=0.$ The two intersect when, say, $t=-(a+b),$ $s=(a+c),$ implying that the point of intersection is $(0, b(a+b), -c(a+c)).$ The other two intersections are $(a(a+b), 0, -c(b+c))$ and $(a(a+c), -b(b+c), 0).$ We need to write a determinant to verify that the three points are collinear:

The right determinant vanishes as the difference of two terms equal to $(a+b)(b+c)(a+c).$ This confirms that the three points are collinear and that $A_b,$ $A_c,$ $B_c,$ $B_a,$ $C_a,$ $C_b$ indeed lie on a conic.